4 5 MOMENT ABOUT AN AXIS Scalar analysis

4. 5 MOMENT ABOUT AN AXIS - Scalar analysis Recall that the moment of a force about any point A is MA= F d. A where d. A is the perpendicular (or shortest) distance from the point to the force’s line of action. This concept can be extended to find the moment of a force about an axis. In the figure above, the moment about the y-axis would be My= 20 (0. 3) = 6 N·m. However, this calculation is not always trivial and vector analysis may be preferable.

VECTOR ANALYSIS Our goal is to find the moment of F (the tendency to rotate the body) about the axis a’-a. First compute the moment of F about any arbitrary point O that lies on the a’a axis using the cross product. MO = r F Now, find the component of MO along the axis a’-a using the dot product. Ma = ua • MO

VECTOR ANALYSIS (continued) Ma can also be obtained as The above equation is also called the triple scalar product. In the this equation, ua represents the unit vector along the axis a’-a axis, r is the position vector from any point on the a’-a axis to any point A on the line of action of the force, and F is the force vector.

EXAMPLE Given: A force is applied to the tool to open a gas valve. A B Find: The magnitude of the moment of this force about the z axis of the value. Plan: 1) We need to use Mz = u • (r F). 2) Note that u = 1 k. 3) The vector r is the position vector from A to B. 4) Force F is already given in Cartesian vector form.

EXAMPLE (continued) u = 1 k r. AB = {0. 25 sin 30° i + 0. 25 cos 30° j+0. 4 k} m B A = {0. 125 i + 0. 2165 j+0. 4 k} m F = {-60 i + 20 j + 15 k} N Mz = u • (r. AB F) Mz = 0 0 0. 125 0. 2165. 125 -60. 216 20 -60 20 1 1 0 0. 4 15 15 = 1{0. 125(20) – 0. 2165(-60)} N·m = 15. 5 N·m

GROUP PROBLEM SOLVING Given: A force of 80 lb acts along the edge DB. Find: The magnitude of the moment of this force about the axis AC. Plan: 1) We need to use M AC = u. AC • (r. AB FDB) 2) Find u. AC = r. AC / r AC 3) Find FDB = 80 lb u. DB = 80 lb (r. DB / r. DB) 4) Complete the triple scalar product.

SOLUTION r. AB = { 20 j } ft r. AC = { 13 i + 16 j } ft r. DB = { -5 i + 10 j – 15 k } ft u. AC = ( 13 i + 16 j ) ft / (13 2 + 16 2 ) ½ ft = 0. 6306 i + 0. 7761 j FDB = 80 {r. DB / (5 2 + 10 2 + 15 2) ½ } lb = {-21. 38 i + 42. 76 j – 64. 14 k } lb

Solution (continued) Now find the triple product, MAC = u. AC • ( r. AB FDB ) MAC = ft lb MAC = 0. 6306 {20 (-64. 14) – 0. 7706 (0 – 0)} lb·ft = -809 lb·ft The negative sign indicates that the sense of MAC is opposite to that of u. AC